But [tex] 0 < A < 1 [/tex], so that is a contradiction and so by theorem 0 < B < A.

I'm not very good at writing proofs down so if anyone has any tips for me or advice on what I can improve on that would be nice. Also I've never used this site before, so if anyone could give me advice on the etiquette of my first post that would be nice. Thanks :)

This shows: [tex]0 < A < 1 [/tex].
This proof shows that A can be defined in complex space and will have same define area as in real.
I think this is all you want if not send me email. I hope I helped you.

I was wondering if someone could further Mr Snoopy's explanation on why his proof is valid. I don't understand why proving it for A as an element of the complex numbers proves it for A as an element of the real numbers.

I tried proving this by assuming the "If" and then manipulating it to no avail.
I then tried proving it by contradiction and I was hoping someone could check my work.

First I negate the question so it is now(If A then B becomes: A and not B):

Before negating, lets look at the statement logically..
Decompose the compound statement into simple ones..
p: 0<B<A <==>(0<B) AND (B<A)
The equivalence of both the statements can be easily verified...

By De Morgan's Rule, [tex]p^{c}[/tex] is [tex](0<B)^{c}[/tex] OR [tex](B<A)^{c}[/tex]..
So the negation would look like: (B<0) OR (A<B) where we are using the inclusive OR...